Answer: +2.10V
Explanation:
[tex]2Al(s)+3I_2(s)\rightarrow 2Al^{3+}(aq)+6I^-(aq)[/tex]
Using Nernst equation :
[tex]E_{cell}=E^o_{cell}-\frac{0.059}{n}\log K[/tex]
[tex]E_{cell}=E^o_{cell}-\frac{0.059}{n}\log [Al^{3+}]^2\times [I^-]^6[/tex]
where,
[tex]E^o_{cell}[/tex] = standard emf for the cell = +2.20 V
n = number of electrons in oxidation-reduction reaction = 6
[tex]E_{cell}[/tex] = emf of the cell = ?
[tex][Al^{3+}][/tex] = concentration = [tex]5.0\times 10^{-3}M[/tex]
[tex][I}^{-}][/tex] = concentration = [tex]0.10M[/tex]
Now put all the given values in the above equation, we get:
[tex]E_{cell}=+2.20-\frac{0.059}{6}\log [5.0\times 10^{-3}]^2\times [0.10]^6[/tex]
[tex]E_{cell}=2.10V[/tex]
The standard emf for the cell using the overall cell reaction below is +2.10 V
